This question is about intersection theory on the easiest (arithmetic) surface over $\mathbf{Z}$: $\mathbf{P}^1_{\mathbf{Z}}$.

Suppose we are given two distinct $\mathbf{Q}$-rational points $b_1$ and $b_2$ on the projective line $\mathbf{P}^1_{\mathbf{Q}}$. Let $P_1$ and $P_2$ be the natural sections of $P^1_{\mathbf{Z}}$ over $\mathbf{Z}$ corresponding to $b_1$ and $b_2$ given by the valuative criterion of properness. (The image of $P_1$ being just the closure of $b_1$ in $\mathbf{P}^1_{\mathbf{Z}}$.)

Can one determine if $P_1\cap P_2 = \emptyset$ just by looking at the generic fibre, i.e., at the points $b_1$ and $b_2$?

Example. The pairs $(b_1,b_2)$ given by $(0,1)$, $(0,\infty)$ or $(1,\infty)$ give sections such that $P_1\cap P_2=\emptyset$. (I can prove this by drawing a picture.)

What am I missing? I'm guessing it could happen that $P_1\cap P_2 =\emptyset$ after composing with an automorphism...

1 Answer
1

Every rational point $b$ of ${\bf P}^1_{\mathbb Q}$ can be written in homogeneous coordinates as $(u : v)$ with coprime integers $u, v\in \mathbb Z$. The vector $(u, v)$ is unique up to signs. The closure $P$ of $b$ has integral homogeneous coordinates $(u:v)$ and the image of $P$ in ${\bf P}^1_{\mathbb F_p}$ has homogeneous coordinates $(\bar{u}: \bar{v})$ with obvious meaning of $\bar{u}$ and $\bar{v}$.

Now write $b_1=(u_1:v_1)$, $b_2=(u_2:v_2)$ as above. Then $P_1$ meets $P_2$ above a prime number $p$ if and only if $(\bar{u}_1:\bar{v}_1)=(\bar{u}_2:\bar{v}_2)$ in ${\bf P}^1$ over ${\mathbb F}_p$. Or equivalently if $u_1v_2-v_1u_2 \equiv 0 \mod p$. Therefore $P_1\cap P_2=\emptyset$ if and only if $u_1v_2-v_1u_2=\pm 1$.

<EDIT> In general the intersection number $(P_1,P_2)$ as defiend in this post is given by
$$ (P_1, P_2)=\log |u_1v_2-v_1u_2|.$$
Actually the intersection number is computed locally at every $p$. At a $p$ prime to $v_1v_2$, the coordinates of $P_i$ is $(u_i/v_i : 1)$ (with $u_i/v_i\in\mathbb Z_p$), so $i_p(P_1, P_2)=v_p(u_1/v_1 - u_2/v_2)=v_p(u_1v_2-u_2v_1)$. If $p$ divides $v_1$ and $P_1, P_2$ intersect each other above $p$, then $p$ divides also $v_2$. Then we do the same computations with $v_i/u_i$ instead of $u_i/v_i$.

A more abstract characterization is, as you suggest, there exists an automorphism of ${\bf P}^1_{\mathbb Z}$ which maps $P_1$ to $(0:1)$ and $P_2$ to $(1:0)$. This is obviously sufficient. The above descriptions shows that if $P_1$ doesn't meet $P_2$, then the automorphism $(x:y)\mapsto (v_1x-u_1y : v_2x-u_2y)$ of ${\bf P}^1_{\mathbb Z}$ maps $b_1$ to $(0:1)$ and $b_2$ to $(1:0)$.

Note that one can't expect a characterization uniquely in terms of the generic fiber because the condition on $P_1, P_2$ depends on a choice of the model ${\bf P}^1_{\mathbb Z}$. For example, $\mathrm{Proj\ } \mathbb Z[2x,y]$ is isomorphic to ${\bf P}^1_{\mathbb Z}$, the points $(1:1), (0:1)$ (in the coordinates $x,y$) meets, in this new model, above $p=2$, but they don't meet in $\mathrm{Proj\ } \mathbb Z[x,y]$.